Question

Formatted question description: https://leetcode.ca/all/199.html

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Algorithm

Just save the rightmost node of each level.

Code

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class Binary_Tree_Right_Side_View {

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<Integer> rightSideView(TreeNode root) {

            List<Integer> result = new ArrayList<>();

            if (root == null) {
                return result;
            }

            Queue<TreeNode> q = new LinkedList<>();
            q.offer(root);

            int currentLevelCount = 1;
            int nextLevelCount = 0;

            while (!q.isEmpty()) {
                TreeNode current = q.poll();
                currentLevelCount--;

                if (current.left != null) {
                    q.offer(current.left);
                    nextLevelCount++;
                }
                if (current.right != null) {
                    q.offer(current.right);
                    nextLevelCount++;
                }

                if (currentLevelCount == 0) {
                    result.add(current.val);
                    currentLevelCount = nextLevelCount;
                    nextLevelCount = 0;
                }
            }

            return result;
        }
    }
}

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            ans.add(q.peekLast().val);
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return ans;
    }
}

// OJ: https://leetcode.com/problems/binary-tree-right-side-view/
// Time: O(N)
// Space: O(H)
class Solution {
    vector<int> ans;
    void dfs(TreeNode *root, int lv) {
        if (!root) return;
        if (lv == ans.size()) ans.push_back(root->val);
        else ans[lv] = root->val;
        dfs(root->left, lv + 1);
        dfs(root->right, lv + 1);
    }
public:
    vector<int> rightSideView(TreeNode* root) {
        dfs(root, 0);
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            ans.append(q[-1].val)
            for _ in range(len(q)):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return ans

#############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque


class Solution(object):
  def rightSideView(self, root):
    """
    :type root: TreeNode
    :rtype: List[int]
    """

    def dfs(root, h):
      if root:
        if h == len(ans):
          ans.append(root.val)
          # pre-order, all the way to the right
        dfs(root.right, h + 1)
        dfs(root.left, h + 1)

    ans = []
    dfs(root, 0)
    return ans

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func rightSideView(root *TreeNode) (ans []int) {
	if root == nil {
		return
	}
	q := []*TreeNode{root}
	for len(q) > 0 {
		ans = append(ans, q[len(q)-1].Val)
		for n := len(q); n > 0; n-- {
			node := q[0]
			q = q[1:]
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	return
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function rightSideView(root: TreeNode | null): number[] {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    while (q.length) {
        const n = q.length;
        ans.push(q[n - 1].val);
        for (let i = 0; i < n; ++i) {
            const { left, right } = q.shift();
            left && q.push(left);
            right && q.push(right);
        }
    }
    return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
    pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = vec![];
        if root.is_none() {
            return res;
        }
        let mut q = VecDeque::new();
        q.push_back(root);
        while !q.is_empty() {
            let n = q.len();
            res.push(q[n - 1].as_ref().unwrap().borrow().val);
            for _ in 0..n {
                if let Some(node) = q.pop_front().unwrap() {
                    let mut node = node.borrow_mut();
                    if node.left.is_some() {
                        q.push_back(node.left.take());
                    }
                    if node.right.is_some() {
                        q.push_back(node.right.take());
                    }
                }
            }
        }
        res
    }
}

199 - Binary Tree Right Side View

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